Suppose that $f :  \Gamma \to \mathbf{R}$ is a trivial quasi-isomorphism and write $f = \phi + g$ where $\phi \in Hom(\Gamma,\mathbf{R})$ and $g : \Gamma \to \mathbf{R}$ is a bounded function as you did. Here we prove that if 
$$|f(x) + f(y) - f(xy)| \le C,$$
for all $x,y \in \Gamma$, then $g$ is bounded by $C$.



Suppose that $g$ is bounded by $D$ and that the bound is optimal, namely $D = \sup_{x \in \Gamma} |g(x)|$. For simplicity, let us assume for the moment that there exists $x \in \Gamma$ such that $|g(x)| = D$. First of all, as $\phi$ is a group homomorphism, we have
$$|g(x) + g(y) - g(xy)|= |f(x) + f(y) - f(xy)|.$$

Since

$$2D - |g(x^2)| \le\left|2|g(x)| - |g(x^2)|\right| \le|2g(x) - g(x^2)| = |2f(x) - f(x^2)| \le C,$$     
we have $|g(x^2)| \ge 2D - C$. On the other hand $D \ge |g(x^2)|$ by assumption, so $C \ge D$. Therefore $g$ is bounded by $C$.      

In the general case, that is if we don't assume that $|g(x)| = D$ for some $x \in \Gamma$, then for every $\varepsilon > 0$ we can still find $x \in \Gamma$ such that $|g(x)| \ge D - \varepsilon$. A similar argument shows that $C +  2\varepsilon \ge D$ for every $\varepsilon > 0$. Therefore $C$ is a bound of $g$.